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insens350 [35]
3 years ago
8

Can someone help me please

Chemistry
1 answer:
MArishka [77]3 years ago
3 0

Answer:

Group 4A (or IVA) of the periodic table includes the nonmetal carbon (C), the metalloids silicon (Si) and germanium (Ge), the metals tin (Sn) and lead (Pb), and the yet-unnamed artificially-produced element ununquadium (Uuq).

The Group 4A elements have four valence electrons in their highest-energy orbitals (ns2np2). Carbon and silicon can form ionic compounds by gaining four electrons, forming the carbide anion (C4-) and silicide anion (Si4-), but they more frequently form compounds through covalent bonding. Tin and lead can lose either their outermost p electrons to form 2+ charges (Sn2+, the stannous ion, and Pb2+, the plumbous ion) or their outermost s and p electrons to form 4+ charges (Sn4+, the stannic ion, and Pb4+, the plumbic ion).

Carbon (C, Z=6).

Carbon is most familiar as a black solid is graphite, coal, and charcoal, or as the hard, crystalline diamond form. The name is derived from the Latin word for charcoal, carbo. It is found in the Earth's crust at a concentration of 480 ppm, making it the 15th most abundant element. It is found in form of calcium carbonate, CaCO3, in minerals such as limestone, marble, and dolomite (a mixture of calcium and

Explanation:

<em><u>T</u></em><em><u>H</u></em><em><u>I</u></em><em><u>S</u></em><em><u> </u></em><em><u>A</u></em><em><u>L</u></em><em><u>L</u></em><em><u> </u></em><em><u>I</u></em><em><u> </u></em><em><u>K</u></em><em><u>N</u></em><em><u>O</u></em><em><u>W</u></em>

<u>E</u><u>N</u><u>J</u><u>O</u><u>Y</u><u> </u><u>THE</u><em><u> </u></em><em><u>A</u></em><em><u>N</u></em><em><u>S</u></em><em><u>W</u></em><em><u>E</u></em><em><u>R</u></em>

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9. What coefficient of H+ balances the atoms in this half-reaction?<br> H+ + MnO₂ → Mn²+ + H₂O
algol13

Answer:4

Explanation:To balance the equation you need to make the number of each element equivalent in both sides.

To start add a 2 in front of the MnO2 which balances the Mn.

Then balance the oxygen by adding a 4 in front of H20.

The H then needs a 8 as it’s coefficient.

7 0
1 year ago
Consider the titration of 100.0 mL of 0.280 M propanoic acid (Ka = 1.3 ✕ 10−5) with 0.140 M NaOH. Calculate the pH of the result
Murljashka [212]

Answer:

(a) 2.7

(b) 4.44

(c) 4.886

(d) 5.363

(e) 5.570

(f)  12.30

Explanation:

Here we have the titration of a weak acid with the strong base NaOH. So in part (a) simply calculate the pH of a weak acid ; in the other parts we have to consider that a buffer solution will be present after some of the weak acid reacts completely the strong base producing the conjugate base. We may even arrive to the situation in which all of the acid will be just consumed and have only  the weak base present in the solution treating it as the pOH and the pH = 14 -pOH. There is also the possibility that all of the weak base will be consumed and then the NaOH will drive the pH.

Lets call HA propanoic acid and A⁻ its conjugate base,

(a) pH = -log √ (HA) Ka =-log √(0.28 x 1.3 x 10⁻⁵) = 2.7

(b) moles reacted HA = 50 x 10⁻³ L x 0.14 mol/L = 0.007 mol

mol left HA = 0.28 - 0.007 = 0.021

mol A⁻ produced = 0.007

Using the Hasselbalch-Henderson equation for buffer solutions:

pH = pKa + log ((A⁻/)/(HA)) = -log (1.3 x 10⁻⁵) + log (0.007/0.021)= 4.89 + (-0.48) = 4.44

(c) = mol HA reacted = 0.100 L x 0.14 mol/L = 0.014 mol

mol HA left = 0.028 -0.014 = 0.014 mol

mol A⁻ produced = 0.014

pH = -log (1.3 x 10⁻⁵) + log (0.014/0.014) =  4.886

(d) mol HA reacted = 150 x 10⁻³ L  x  x 0.14 mol/L = 0.021 mol

mol HA left = 0.028 - 0.021 = 0.007

mol A⁻ produced = 0.021

pH = -log (1.3 x 10⁻⁵) + log (0.021/0.007) =  5.363

(e) mol HA reacted = 200 x 10⁻³ L x 0.14 mol/L = 0.028 mol

mol HA left = 0

Now we only a weak base present and its pH is given by:

pH  = √(kb x (A⁻)  where Kb= Kw/Ka

Notice that here we will have to calculate the concentration of A⁻ because we have dilution effects the moment we added to the 100 mL of HA,  200 mL of NaOH 0.14 M. (we did not need to concern ourselves before with this since the volumes cancelled each other in the previous formulas)

mol A⁻ = 0.028 mOl

Vol solution = 100 mL + 200 mL = 300 mL

(A⁻) = 0.028 mol /0.3 L = 0.0093 M

and we also need to calculate the Kb for the weak base:

Kw = 10⁻¹⁴ = ka Kb ⇒   Kb = 10⁻¹⁴/1.3x 10⁻⁵ = 7.7 x 10⁻ ¹⁰

pH = -log (√( 7.7 x 10⁻ ¹⁰ x 0.0093) = 5.570

(f) Treat this part as a calculation of the pH of a strong base

moles of OH = 0.250 L x 0.14 mol = 0.0350 mol

mol OH remaining = 0.035 mol - 0.028 reacted with HA

= 0.007 mol

(OH⁻) = 0.007 mol / 0.350 L = 2.00 x 10 ⁻²

pOH = - log (2.00 x 10⁻²) = 1.70

pH = 14 - 1.70 = 12.30

4 0
3 years ago
I need to find the chemical equation
Yuri [45]

Answer:

2NaCN + CaCO3 --> Na2CO3 + Ca(CN)2

Explanation:

Knowing the names gets us: NaCN + CaCO3 --> Na2CO3 + Ca(CN)2

Balance: there are two sodiums and cyanides on the product side so add a 2 to the reactant side.

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Write a balanced equation for the double-replacement precipitation reaction described, using the smallest possible integer coeff
ahrayia [7]

Answer:

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Explanation:

When ammonium bromide and silver(I) nitrate react, the following are obtained as shown below:

NH4Br(aq) + AgNO3(aq) —>

In solution, NH4Br(aq) and AgNO3(aq) will dissociate as follow:

NH4Br(aq) —> NH4+(aq) + Br-(aq)

AgNO3(aq) —> Ag+(aq) + NO3-(aq)

The double displacement reaction will occur as follow:

NH4+(aq) + Br-(aq) + Ag+(aq) + NO3-(aq) —> Ag+(aq) + Br-(aq) + NH4+(aq) + NO3-(aq)

NH4Br(aq) + AgNO3(aq) —> AgBr(s) + NH4NO3(aq)

6 0
3 years ago
How many significant figures are in 3.02 x 108 ?<br> a<br> 2<br> b 4<br> C 5<br> d 3
bija089 [108]

Answer:

C

Explanation:

the answer has 5 Numbers

7 0
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